M216: Exercise sheet 1

Warmup questions

• 1. Let $$U$$ be a subset of a vector space $$V$$. Show that $$U$$ is a linear subspace of $$V$$ if and only if $$U$$ satisfies the following conditions:

• (i) $$0\in U$$;

• (ii) For all $$u_1,u_2\in U$$ and $$\lambda \in \F$$, $$u_1+\lambda u_2\in U$$.

• 2. Which of the following subsets of $$\R ^3$$ are linear subspaces? In each case, briefly justify your answer.

(a) $$U_1:=\set {(x_1,x_2,x_3)\st x_1^2+x_2^2+x_3^2=1}$$ (b) $$U_2:=\set {(x_1,x_2,x_3)\st x_1=x_2}$$ (c) $$U_3:=\set {(x_1,x_2,x_3)\st x_1+2x_2+3x_3=0}$$

• 3. Which of the following maps $$f:\R ^2\to \R ^2$$ are linear? In each case, briefly justify your answer.

(a) $$f(x,y)=(5x+y,3x-2y)$$ (b) $$f(x,y)=(5x+2,7y)$$ (c) $$f(x,y)=(\cos y,\sin x)$$ (d) $$f(x,y)=(3y^{2},x^3)$$.

Homework

• 4. Let $$\cI$$ be a set and $$V$$ a vector space over a field $$\F$$. Recall that $$V^{\cI }$$ is the set of maps $$\cI \to V$$.

Show that $$V^{\cI }$$ is a vector space under pointwise addition and scalar multiplication.

• 5. Let $$\R [x]$$ be the space of real polynomials. This is a vector space under coefficient-wise addition and scalar multiplication.

For $$d\in \N$$, let $$P_{d}\subset \R [x]$$ be the set of polynomials of degree no more than $$d$$. Show that $$P_d\leq \R [x]$$ and has basis $$1,x,\dots , x^{d}$$

Define a linear map $$D:P_d\to P_d$$ by $$D(p)=p'$$. Compute its matrix with respect to $$1,x,\dots , x^{d}$$. What are $$\ker D$$ and $$\im D$$?

• 6. Which of the following subsets of $$\C ^3$$ are linear subspaces over $$\C$$? In each case, briefly justify your answer.

(a) $$U_1:=\set {(z_1,z_2,z_3)\st z_1z_2=1}$$ (b) $$U_2:=\set {(z_1,z_2,z_3)\st z_1=\bar {z}_2}$$ (c) $$U_3:=\set {(z_1,z_2,z_3)\st z_1+\sqrt {-1}z_2+3z_3=0}$$

• 7. Let $$V$$ be an $$n$$-dimensional vector space over $$\C$$, and let $$V_\R$$ be the underlying vector space over $$\R$$ (thus $$V_\R$$ has the same set of vectors as $$V$$, but scalar multiplication is restricted to real scalars). Prove that $$V_\R$$ has dimension $$2n$$.

[Hint: let $$\cB :v_1,v_2,\ldots ,v_n$$ be a basis for $$V$$ and show that $$\cB _\R :v_1,i v_1, v_2,i v_2, \ldots ,v_n,i v_n$$ is a basis for $$V_\R$$, where $$i\in \C$$ is $$\sqrt {-1}$$ rather than an index!]

Please hand in at 4W level 1 by NOON on Friday 13th October 2023

M216: Exercise sheet 1—Solutions

• 1. First suppose that $$U\leq V$$. The $$U$$ is non-empty so there is some $$u\in U$$ and then, since $$U$$ is closed under addition and scalar multiplication, $$0=u+(-1)u\in U$$ also and condition (i) is satisfied. Now if $$u_1, u_2\in U$$ and $$\lambda \in \F$$, then $$\lambda u_2\in U$$ ($$U$$ is closed under scalar multiplication) and so $$u_1+\lambda u_2\in U$$ ($$U$$ is closed under addition). Thus condition (ii) holds also.

For the converse, if conditions (i) and (ii) hold, then, first, $$0\in U$$ so $$U$$ is non-empty and, second, $$U$$ is closed under addition (take $$\lambda =1$$ in condition (ii)) and under scalar multiplication (take $$u_1=0$$ in condition (ii)). Thus $$U\leq V$$.

• 2.

• (a) $$U_1$$ is not a subspace as it does not contain $$0$$!

• (b) $$U_2$$ is a subspace: in fact, it is $$\ker \phi _A$$ where $$A= \begin {pmatrix} 1&-1&0 \end {pmatrix}$$.

• (c) $$U_3$$ is a subspace. It is $$\ker \phi _A$$ for $$A= \begin {pmatrix} 1&2&3 \end {pmatrix}$$.

• 3.

• (a) Here $$f$$ is linear: it is the map $$\phi _A$$ corresponding to the matrix

\begin{equation*} A= \begin{pmatrix} 5&1\\3&-2 \end {pmatrix}. \end{equation*}

• (b) This is not linear (because of that $$+2$$ term). In particular $$f(0,0)=(2,0)\neq 0$$!

• (c) Again $$f(0,0)=(1,0)\neq 0$$ so this $$f$$ cannot be linear. Of course, we already know this because it is certainly not true that $$\cos (y_1+y_2)=\cos y_1+\cos y_2$$.

• (d) Another non-linear map: for example $$f(2x,2y)\neq 2f(x,y)$$.

• 4. The basic idea is that the vector space axioms for $$V^{\cI }$$ will follow from those of $$V$$ applied to the values of elements of $$V^{\cI }$$. Since those elements are completely determined by their values, this will bake the cake.

In more detail: let $$u,v,w\in V^{\cI }$$, then, for $$i\in \cI$$,

\begin{equation*} (u+v)(i)=u(i)+v(i)=v(i)+u(i)=(v+u)(i), \end{equation*}

whence $$u+v=v+u$$. Here the first and last equalities are just the definition of pointwise addition and the middle one of commutativity of addition in $$V$$.

Similarly,

\begin{equation*} ((u+v)+w)(i)=(u+v)(i)+w(i)=(u(i)+v(i))+w(i)= u(i)+(v(i)+w(i))=(u+(v+w))(i) \end{equation*}

so that $$(u+v)+w=u+(v+w)$$.

The zero element is the zero map defined by $$0(i):=0$$, for all $$i\in \cI$$, while the additive inverse $$-v$$ of $$v\in V^{\cI }$$ is defined by $$(-v)(i):=-(v(i))$$. Now

\begin{align*} (v+0)(i)&=v(i)+0(i)=v(i)+0=v(i)\\ (v+(-v))(i)&=v(i)+(-v)(i)=v(i)-v(i)=0=0(i) \end{align*} so that $$v+0=v$$ and $$v+(-v)=0$$ as required.

The axioms around scalar multiplication are verified in the same way. For example, for $$\lambda ,\mu \in \F$$,

\begin{equation*} ((\lambda +\mu )v)(i)=(\lambda +\mu )(v(i))=\lambda (v(i))+\mu (v(i))=(\lambda v)(i)+(\mu v)(i)=(\lambda v+\mu v)(i) \end{equation*}

so that $$(\lambda +\mu )v=\lambda v+\mu v$$.

Again, for $$u,v\in V^{\cI }$$ and $$\lambda \in \F$$,

\begin{multline*} (\lambda (u+v))(i)=\lambda (u+v)(i)=\lambda (u(i)+v(i))=\lambda u(i)+\lambda v(i)\\ =(\lambda u)(i)+(\lambda v)(i)=(\lambda u+\lambda v)(i) \end{multline*} so that $$\lambda (u+v)=\lambda u+\lambda v$$.

For $$\lambda ,\mu \in F$$ and $$v\in V^{\cI }$$,

\begin{equation*} ((\lambda \mu )v)(i)=(\lambda \mu )v(i)=\lambda (\mu v(i))=(\lambda (\mu v))(i) \end{equation*}

so that $$(\lambda \mu )v=\lambda (\mu v)$$.

Finally, $$(1v)(i)=1v(i)=v(i)$$ so that $$1v=v$$ and we are (at last!) done.

• 5. Clearly $$P_d$$ is non-empty as it contains the zero polynomial. Moreover, for any polynomials $$p,q$$ and $$\lambda \in \R$$, we have

\begin{align*} \deg (p+q)&\leq \max \set {\deg p,\deg q}\\ \deg (\lambda p)&\leq \deg p, \end{align*} from which it easily follows that $$P_d$$ is closed under addition and scalar multiplication.

Any polynomial $$p\in P_d$$ has a unique expression of the form

\begin{equation*} p=a_0+a_1x+\dots +a_dx^{d}. \end{equation*}

It now follows from Proposition 1.1 that $$1,\rng {x}{x^d}$$ is a basis for $$P_d$$.

Set $$v_j=x^{j-1}$$, for $$1\leq j\leq d+1$$, and compute $$D v_j$$ in terms of the $$v_i$$:

\begin{equation*} D v_j=(j-1)v_{j-1} \end{equation*}

so that the matrix $$A$$ of $$D$$ with respect to this basis has all entries 0 except just above the diagonal where $$A_{(j-1)j}=j-1$$. For example, if $$d=3$$, we have

\begin{equation*} A= \begin{pmatrix} 0&1&0&0\\0&0&2&0\\0&0&0&3\\0&0&0&0 \end {pmatrix}. \end{equation*}

The kernel of $$D$$ is the constant polynomials $$P_0$$ and the image is $$P_{d-1}$$.

• 6.

• (a) $$0\notin U_1$$ so $$U_1$$ is not a subspace.

• (b) $$U_2$$ is not a subspace because it is not closed under complex scalar multiplication: $$(1,1,0)\in U_2$$ but $$i(1,1,0)=(i,i,0)$$ is not (here $$i=\sqrt {-1}$$). In general, any time you see complex conjugation in the definition of a subset, it is unlikely to be a complex subspace.

• (c) $$U_3=\ker \phi _A$$ for $$A= \begin {pmatrix} 1&\sqrt {-1}&3 \end {pmatrix}$$ and so is a subspace.

• 7. Following the hint we need to show that any $$v\in V_\R$$ can be written uniquely as a real linear combination of vectors in the list $$\cB _\R$$. Since $$v\in V$$, we may write $$v=\sum _{j=1}^n \lambda _j v_j$$ for unique $$\lambda _j\in \C$$. Write $$\lambda _j=a_j+i b_j$$ with $$a_j,b_j\in \R$$. Then $$v=\sum _{j=1}^n (a_j v_j+b_j i v_j)$$ and this expression is unique: it suffices to observe that for $$v=0$$, $$\lambda _j=0$$ for all $$j$$, and hence $$a_j=b_j=0$$ for all $$j$$.